Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include <cstdio>
#include <map>
using namespace std;
typedef long long LL;
const int N=250010;
const LL MOD=1000000000000000031ll,B=1000000000000000003ll,REVB=821428571428571454ll;
int n;
char str[N];
LL f[N],p[N];
LL mul(LL x,LL y){
	return (LL)((__int128)x*y%MOD);
}
void readData(){
	scanf("%d",&n);
	scanf("%s",str+1);
}
map<LL,int> mp;
void solve(){
	f[0]=0;
	p[0]=1;
	for(int i=1;i<=n;i++){
		f[i]=f[i-1];
		p[i]=p[i-1];
		if(str[i]=='>')
			p[i]=mul(p[i],B);
		else if(str[i]=='<')
			p[i]=mul(p[i],REVB);
		else if(str[i]=='+')
			(f[i]+=p[i])%=MOD;
		else
			(f[i]-=p[i])%=MOD;
		if(f[i]<0)
			f[i]+=MOD;
	}
	LL ans=0;
	for(int i=n;i>=1;i--){
		mp[f[i]]++;
		ans+=mp[(mul(f[n],p[i-1])+f[i-1])%MOD];
	}
	printf("%lld\n",ans);
}
int main(){
	readData();
	solve();
	return 0;
}